1. Connected Components, Directed graphs, Topological sort COMP171 Fall 2005

2. Graph / Slide 2 Application 1: Connectivity D E A C F B G K H L N M O R Q P s How do we tell if two vertices are connected? A connected to F? A connected to L? G =

3. Graph / Slide 3 Connectivity * A graph is connected if and only if there exists a path between every pair of distinct vertices. * A graph is connected if and only if there exists a simple path between every pair of distinct vertices (since every non-simple path contains a cycle, which can be bypassed) * How to check for connectivity? n Run BFS or DFS (using an arbitrary vertex as the source) n If all vertices have been visited, the graph is connected. n Running time? O(n + m)

4. Graph / Slide 4 Connected Components

5. Graph / Slide 5 Subgraphs

6. Graph / Slide 6 Connected Components * Formally stated: n A connected component is a maximal connected subgraph of a graph. n The set of connected components is unique for a given graph.

7. Graph / Slide 7 Finding Connected Components For each vertex Call DFS This will find all vertices connected to “v” => one connected component Basic DFS algorithm. If not visited

8. Graph / Slide 8 Time Complexity * Running time for each i connected component * Question: n Can two connected components have the same edge? n Can two connected components have the same vertex? * So:

9. Graph / Slide 9 Trees * Tree arises in many computer science applications * A graph G is a tree if and only if it is connected and acyclic (Acyclic means it does not contain any simple cycles) * The following statements are equivalent n G is a tree n G is acyclic and has exactly n-1 edges n G is connected and has exactly n-1 edges

10. Graph / Slide 10 Tree as a (directed) Graph 15 618 3 8 30 16 Is it a graph? Does it contain cycles? (in other words, is it acyclic) How many vertices? How many edges?

11. Graph / Slide 11 Directed Graph * A graph is directed if direction is assigned to each edge. We call the directed edges arcs. n An edge is denoted as an ordered pair (u, v) * Recall: for an undirected graph n An edge is denoted {u,v}, which actually corresponds to two arcs (u,v) and (v,u)

12. Graph / Slide 12 Representations * The adjacency matrix and adjacency list can be used

13. Graph / Slide 13 Directed Acyclic Graph * A directed path is a sequence of vertices (v 0, v 1,..., v k ) n Such that (v i, v i+1 ) is an arc * A directed cycle is a directed path such that the first and last vertices are the same. * A directed graph is acyclic if it does not contain any directed cycles

14. Graph / Slide 14 Indegree and Outdegree * Since the edges are directed n We can’t simply talk about Deg(v) * Instead, we need to consider the arcs coming “in” and going “out” n Thus, we define terms  Indegree(v)  Outdegree(v)

15. Graph / Slide 15 Outdegree * All of the arcs going “out” from v * Simple to compute n Scan through list Adj[v] and count the arcs * What is the total outdegree? (m=#edges)

16. Graph / Slide 16 Indegree * All of the arcs coming “in” to v * Not as simple to compute as outdegree n First, initialize indegree[v]=0 for each vertex v n Scan through adj[v] list for each v  For each vertex w seen, indegree[w]++;  Running time: O(n+m) * What is the total indegree?

17. Graph / Slide 17 Indegree + Outdegree * Each arc (u,v) contributes count 1 to the outdegree of u and count 1 to the indegree of v.

18. Graph / Slide 18 Example 0 1 2 3 4 5 6 7 8 9 Indeg(2)? Indeg(8)? Outdeg(0)? Num of Edges? Total OutDeg? Total Indeg?

19. Graph / Slide 19 Directed Graphs Usage * Directed graphs are often used to represent order- dependent tasks * That is we cannot start a task before another task finishes * We can model this task dependent constraint using arcs * An arc (i,j) means task j cannot start until task i is finished * Clearly, for the system not to hang, the graph must be acyclic. i j Task j cannot start until task i is finished

20. Graph / Slide 20 University Example * CS departments course structure 171151 180 211251 272 271 252 M132M111 231 201 221 361 362 381 303 327 336 341 343 342 332 334 104 How many indeg(171)? How many outdeg(171)? Any directed cycles?

21. Graph / Slide 21 Topological Sort * Topological sort is an algorithm for a directed acyclic graph * It can be thought of as a way to linearly order the vertices so that the linear order respects the ordering relations implied by the arcs 0 1 2 3 4 5 6 7 8 9 For example: 0, 1, 2, 5, 9 0, 4, 5, 9 0, 6, 3, 7 ?

22. Graph / Slide 22 Topological Sort * Idea: n Starting point must have zero indegree! n If it doesn’t exist, the graph would not be acyclic 1. A vertex with zero indegree is a task that can start right away. So we can output it first in the linear order 2. If a vertex i is output, then its outgoing arcs (i, j) are no longer useful, since tasks j does not need to wait for i anymore- so remove all i’s outgoing arcs 3. With vertex i removed, the new graph is still a directed acyclic graph. So, repeat step 1-2 until no vertex is left.

23. Graph / Slide 23 Topological Sort Find all starting points Reduce indegree(w) Place new start vertices on the Q

24. Graph / Slide 24 Example 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 1 2 1 1 2 1 1 2 2 Indegree start Q = { 0 } OUTPUT: 0

25. Graph / Slide 25 Example 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 1 2 1 1 2 1 1 2 2 Indegree Dequeue 0 Q = { } -> remove 0’s arcs – adjust indegrees of neighbors OUTPUT: Decrement 0’s neighbors

26. Graph / Slide 26 Example 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 2 1 0 2 0 1 2 2 Indegree Dequeue 0 Q = { 6, 1, 4 } Enqueue all starting points OUTPUT: 0 Enqueue all new start points

27. Graph / Slide 27 Example 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 2 1 0 2 0 1 2 2 Indegree Dequeue 6 Q = { 1, 4 } Remove arcs.. Adjust indegrees of neighbors OUTPUT: 0 6 Adjust neighbors indegree

28. Graph / Slide 28 Example 1 2 3 4 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 1 0 0 2 0 1 2 2 Indegree Dequeue 6 Q = { 1, 4, 3 } Enqueue 3 OUTPUT: 0 6 Enqueue new start

29. Graph / Slide 29 Example 1 2 3 4 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 1 0 0 2 0 1 2 2 Indegree Dequeue 1 Q = { 4, 3 } Adjust indegrees of neighbors OUTPUT: 0 6 1 Adjust neighbors of 1

30. Graph / Slide 30 Example 2 3 4 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 2 0 1 2 2 Indegree Dequeue 1 Q = { 4, 3, 2 } Enqueue 2 OUTPUT: 0 6 1 Enqueue new starting points

31. Graph / Slide 31 Example 2 3 4 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 2 0 1 2 2 Indegree Dequeue 4 Q = { 3, 2 } Adjust indegrees of neighbors OUTPUT: 0 6 1 4 Adjust 4’s neighbors

32. Graph / Slide 32 Example 2 3 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 1 0 1 2 2 Indegree Dequeue 4 Q = { 3, 2 } No new start points found OUTPUT: 0 6 1 4 NO new start points

33. Graph / Slide 33 Example 2 3 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 1 0 1 2 2 Indegree Dequeue 3 Q = { 2 } Adjust 3’s neighbors OUTPUT: 0 6 1 4 3

34. Graph / Slide 34 Example 2 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 1 0 1 1 2 Indegree Dequeue 3 Q = { 2 } No new start points found OUTPUT: 0 6 1 4 3

35. Graph / Slide 35 Example 2 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 1 0 1 1 2 Indegree Dequeue 2 Q = { } Adjust 2’s neighbors OUTPUT: 0 6 1 4 3 2

36. Graph / Slide 36 Example 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 1 2 Indegree Dequeue 2 Q = { 5, 7 } Enqueue 5, 7 OUTPUT: 0 6 1 4 3 2

37. Graph / Slide 37 Example 5 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 1 2 Indegree Dequeue 5 Q = { 7 } Adjust neighbors OUTPUT: 0 6 1 4 3 2 5

38. Graph / Slide 38 Example 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 1 1 Indegree Dequeue 5 Q = { 7 } No new starts OUTPUT: 0 6 1 4 3 2 5

39. Graph / Slide 39 Example 7 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 1 1 Indegree Dequeue 7 Q = { } Adjust neighbors OUTPUT: 0 6 1 4 3 2 5 7

40. Graph / Slide 40 Example 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 1 Indegree Dequeue 7 Q = { 8 } Enqueue 8 OUTPUT: 0 6 1 4 3 2 5 7

41. Graph / Slide 41 Example 8 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 1 Indegree Dequeue 8 Q = { } Adjust indegrees of neighbors OUTPUT: 0 6 1 4 3 2 5 7 8

42. Graph / Slide 42 Example 9 0 1 2 3 4 5 6 7 8 9 2 614 75 8 5 32 8 9 9 0 1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 0 0 Indegree Dequeue 8 Q = { 9 } Enqueue 9 Dequeue 9 Q = { } STOP – no neighbors OUTPUT: 0 6 1 4 3 2 5 7 8 9

43. Graph / Slide 43 Example OUTPUT: 0 6 1 4 3 2 5 7 8 9 0 1 2 3 4 5 6 7 8 9 Is output topologically correct?

44. Graph / Slide 44 Topological Sort: Complexity * We never visited a vertex more than one time * For each vertex, we had to examine all outgoing edges n Σ outdegree(v) = m n This is summed over all vertices, not per vertex * So, our running time is exactly n O(n + m)